🚀 go-pugleaf

In article <1992Aug19.172922.410@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>>	Kemeny's conjecture is false: If M is a non-standard model of
>>PA, then there is a non-standard integer n such that no integer in the
>>"standard row" of n is divisible by all standard integers.
>>
>>	To see this, take n which is odd but is divisible by all
>>standard odd integers. If m differs from n by a standard integer, then
>>there must be a standard odd prime which does not divide the
>>difference. Hence m is not divisible by that prime.

>
>Looking at this bit...
>

>>take n which is odd but is divisible by all standard odd integers.

>
>...at first it wasn't clear to me that every nonstandard model would have
>such an  n .  But I suppose this can be seen by taking any odd nonstandard k,
>and forming  n = k(k-2)(k-4)(k-6).....  .

If what is meant here is that n equals to the product of the odd predecessors
in the _row_ of k, then there is no model with such n, just as there is no
number equal to the product of all standard integers; the reason for that is
overspill, or else that by being able to do that you could pin down the
standard integers.(Boolos&Jeffry surely discusses that).
To see that there is an odd non-standard divisible by all odd standard naturals
just take any non-standard that is divisible by all standard numbers and get
its greatest odd divisor. You can prove that there is always such a divisor in
PA for all numbers, therefore it is true in all models of PA.

>
>Presumably this operation can be made legitimate in some simple way ?

Yes, by Godel coding, but n will necessarily be divisible by some non-standard
naturals.
WJCG